question_answer

Consider the charge q, q and -q placed at the vertices of an equilateral triangle of each side L. The sum of forces acting on each charge is:

A) \[\frac{{{q}^{2}}}{4\sqrt{2}\pi {{\varepsilon }_{0}}{{l}^{2}}}\]                     

B) \[\frac{-{{q}^{2}}}{4\pi {{\varepsilon }_{0}}{{l}^{2}}}\]

C) \[\frac{{{q}^{2}}}{4\pi {{\varepsilon }_{0}}{{l}^{2}}}\]                    

D) zero

Correct Answer: D

Solution :

(d) zero From diagram, force on\[{{q}_{1}}\left( =q \right)\] at A, \[{{\overrightarrow{F}}_{1}}{{\overrightarrow{F}}_{12}}+{{\overrightarrow{F}}_{13}}=F{{\hat{r}}_{1}}\] where,  \[F=\frac{{{q}^{2}}}{4\pi {{\varepsilon }_{0}}{{l}^{2}}}\]and \[{{\hat{r}}_{1}}\]is the unit vector along B(C) Force on \[{{q}_{2}}\left( =q \right)\]at B; \[{{\overrightarrow{F}}_{2}}={{\overrightarrow{F}}_{21}}+{{\overrightarrow{F}}_{23}}=F{{\hat{r}}_{2}}\] where, \[{{\hat{r}}_{2}}\]is the unit vector along A(C) Force on \[{{q}_{3}}\left( =-q \right)\]at C, \[{{\overrightarrow{F}}_{3}}={{\overrightarrow{F}}_{31}}+{{\overrightarrow{F}}_{32}}\] \[=\left( \sqrt{{{F}^{2}}+{{F}^{2}}+2F\,.\,F\,\cos \,60{}^\circ } \right)\hat{n}=\sqrt{3}\,F\,\hat{n}\] where, \[\hat{n}=unit\]vector along the direction bisecting \[\angle BCA\] \[\therefore \,\,\,\,\,\,\,{{\overrightarrow{F}}_{1}}+{{\overrightarrow{F}}_{2}}+{{\overrightarrow{F}}_{3}}=0\]